# NCERT Solutions Class 8 Mathematics Solutions for Exercise 9.5 in Chapter 9 - Algebraic Expressions and Identities

Question 12 Exercise 9.5

Using identities, evaluate.

\text { (i) } 71^{2}

\text { (ii) } 99^{2}

(iii) 102^{2}

\text { (iv) } 998^{2}

\text { (v) } 5.2^{2}

\text { (vi) } 297 \times 303

\text { (vii) } 78 \times 82

\text { (viii) } 8.9^{2}

\text { (ix) } 10.5 \times 9.5

\begin{aligned} &\text { i) } 71^{2}=(70+1)^{2}\\ &\begin{array}{l} =70^{2}+140+1^{2} \\ =4900+140+1 \\ =5041 \end{array} \end{aligned}

\begin{aligned} &\text { ii) } 99^{2}\\ &\begin{array}{l} =(100-1)^{2} \\ =100^{2}-200+1^{2} \\ =10000-200+1 \\ =9801 \end{array} \end{aligned}

\begin{aligned} &\text { iii) } 102^{2}=(100+2)^{2}\\ &\begin{array}{l} =100^{2}+400+2^{2} \\ =10000+400+4 \\ =10404 \end{array} \end{aligned}

\begin{aligned} &\text { iv) } 998^{2}=(1000-2)^{2}\\ &\begin{array}{l} =1000^{2}-4000+2^{2} \\ =1000000-4000+4 \\ =996004 \end{array} \end{aligned}

\begin{aligned} &\text { v) } 5.2^{2}=(5+0.2)^{2}\\ &\begin{array}{l} =5^{2}+2+0.2^{2} \\ =25+2+0.4 \\ =27.4 \end{array} \end{aligned}

\begin{aligned} &\text { vi) } 297 \times 303\\ &=(300-3)(300+3)\\ &=300^{2}-3^{2}\\ &=90000-9\\ &=89991 \end{aligned}

\begin{aligned} &\text { vii) } 78 \times 82\\ &\begin{array}{l} =(80-2)(80+2) \\ =80^{2}-2^{2} \\ =6400-4 \\ =6396 \end{array} \end{aligned}

\begin{aligned} &\text { viii) } 8.9^{2}=(9-0.1)^{2}\\ &\begin{array}{l} =9^{2}-1.8+0.1^{2} \\ =81-1.8+0.01 \\ =79.21 \end{array} \end{aligned}

\begin{aligned} &\text { ix) } 10.5 \times 9.5\\ &=(10+0.5)(10-0.5)\\ &=10^{2}-0.5^{2}\\ &=100-0.25\\ &=99.75 \end{aligned}

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